822. Card Flipping Game

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822. Card Flipping Game

题目描述

On a table are N cards, with a positive integer printed on the front and back of each card (possibly different).

We flip any number of cards, and after we choose one card.

If the number X on the back of the chosen card is not on the front of any card, then this number X is good.

What is the smallest number that is good? If no number is good, output 0.

Here, fronts[i] and backs[i] represent the number on the front and back of card i.

A flip swaps the front and back numbers, so the value on the front is now on the back and vice versa.

Example:

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Input: fronts = [1,2,4,4,7], backs = [1,3,4,1,3]
Output: 2
Explanation: If we flip the second card, the fronts are [1,3,4,4,7] and the backs are [1,2,4,1,3].
We choose the second card, which has number 2 on the back, and it isn't on the front of any card, so 2 is good.

Note:

  • 1 <= fronts.length == backs.length <= 1000.
  • 1 <= fronts[i] <= 2000.
  • 1 <= backs[i] <= 2000.

题目大意

每张牌的正反面都有一个数，随意翻牌，找出一个最小的数与其他牌当前的正面数值不相等。
front 代表牌正面的数值，back 代表牌背面的数值。

解题思路

如果正反面数值相同，则无论我们如何翻这张牌，正面的牌一定不符合条件。
找出正反面数值相同。将数值存在集合内。
遍历 front 和 back ，找到不在集合中存在的最小数。

代码

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class Solution {
    func flipgame(_ fronts: [Int], _ backs: [Int]) -> Int {
        var same = Set<Int>()
        for (front, back) in zip(fronts, backs) {
            if front == back {
                same.insert(front)
            }
        }
        var ans = Int.max
        for (front, back) in zip(fronts, backs) {
            if !same.contains(front){ ans = min(ans, front) }
            if !same.contains(back) { ans = min(ans, back) }
        }
        return ans == Int.max ? 0 : ans
    }
}
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